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irakobra [83]
3 years ago
13

The concentration of phenylephrine hydrochloride in a prefilled syringe is 10 mg/mL. What is the percentage concentration (w/v)

of phenylephrine in the injection. (Answer must be numeric; no units or commas; round the final answer to the nearest WHOLE number.)
Chemistry
1 answer:
mihalych1998 [28]3 years ago
6 0

Answer:

The percentage concentration (w/v) of phenylephrine in the injection is 1 %.

Explanation:

The percentage concentration in w/v of the phenylephrine hydrochloride can be calculated as follows:

\%^{w}/_{v} = \frac{m}{V} \times 100

Where:

m: is the mass of the solute in grams

V: is the volume of the solution in milliliters

The concentration of the phenylephrine hydrochloride is 10 mg/mL, so the percentage concentration is:

\%^{w}/_{v} = \frac{m}{V} \times 100

\%^{w}/_{v} = \frac{10 \cdot 10^{-3} g}{1 mL} \times 100

\%^{w}/_{v} = 1 \%

Therefore, the percentage concentration (w/v) of phenylephrine in the injection is 1 %.

I hope it helps you!

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Answer:

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3 years ago
1). How many moles of H₂ would be required to produce 9.0 grams of water?
kramer

1. How many moles of H₂ would be required to produce 9.0 grams of water?

The answer would be approximately 0.4495 moles of H₂.

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A mass of  72g H₂O will be formed.

3.How many moles of SnF₂ will be produced along with 48 grams of H₂?

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3 0
4 years ago
A beaker with 1.80×102 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and c
tekilochka [14]

Answer:

The pH change in 0,206 units

Explanation:

When the acetic acid buffer is at pH 5,000; it is possible to obtain the acetate/acetic acid proportions using Henderson-Hasselbalch formula, thus:

pH = pka + log₁₀ [A⁻]/[HA] Where A⁻ is CH₃COO⁻ and HA is CH₃COOH.

Replacing:

5,000 = 4,740 + log₁₀ [A⁻]/[HA]

1,820 = [A⁻]/[HA] <em>(1)</em>

As buffer concentration is 0,100M:

[A⁻] + [HA] = 0,100 <em>(2)</em>

Replacing (2) in (1)

[HA] = 0,035M

And [A⁻] = 0,065M

As volume is 1,80x10²mL, moles of HA and A⁻ are:

0,180L × 0,035M = <em>6,3x10⁻³mol of HA</em>

0,180L × 0,065M = <em>1,17x10⁻²mol of A⁻</em>

The reaction of HCl with A⁻ is:

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The add moles of HCl are:

0,0065L×0,330M = 2,145x10⁻³ moles of HCl that are equivalent to moles of A⁻ consumed and moles of HA produced.

Thus, moles of HA after addition of HCl are:

6,3x10⁻³mol + 2,145x10⁻³ mol = <em>8,445x10⁻³ moles of HA</em>

And moles of A⁻ are:

1,17x10⁻²mol - 2,145x10⁻³ mol = <em>9,555x10⁻³ moles of A⁻</em>

Replacing these values in Henderson-Hasselbalch formula:

pH = 4,740 + log₁₀ [9,555x10⁻³ ]/[8,445x10⁻³ ]

pH = 4,794

<em>The pH change in </em>5,000-4,794 <em>= 0,206 units</em>

I hope it helps!

8 0
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LuckyWell [14K]

Answer:

105

Explanation:

Density = mass/volume

therefore volume = mass/density

826/7.9=105 (To three significant figure)

8 0
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